I'm attempting to create a scoring system for a card game which would preclude ties in scoring, by setting the point value of each card such that no two combinations of cards could add up to the same score. (For this particular case, I need a set of 17 integers, since there are 17 scorable cards.)
I've tried several heuristic approaches (various winnowing procedures along the lines of taking an array of integers, iteratively generating random subsets, and discard those which appear in subsets sharing a common sum); then exhaustively validating the results (by enumerating their subsets).
From what I've seen, the theoretical limit to the size of such a set is near log2(n), where n is the number of members of the superset from which the subset-distinct-sum subset is drawn. However, while I've been able to approach this, I've not been able to match it. My best result so far is a set of 13 integers, drawn from the 250,000 integers between 10,000 and 25,000,000, counting by hundreds (the latter is immaterial to the algorithm, but is a domain constrain of my use case):
[332600,708900,2130500,2435900,5322500,7564200,10594500,12776200,17326700,17925700,22004400,23334700,24764900]
I've hunted around, and most of the SDS generators are sequence generators that make no pretense of creating dense sets, but instead have the ability to be continued indefinitely to larger and larger numbers (e.g. the Conway-Guy Sequence). I have no such constraint, and would prefer a denser set without requiring a sequence relationship with each other.
(I did consider using the Conway-Guy Sequence n=2..18 * 10,000, but the resulting set has a broader range than I would like. I'd also really like a more general algorithm.)
Edit: For clarity, I'm looking for a way (non-deterministic or dynamic-programming methods are fine) to generate an SDS set denser than those provided by simply enumerating exponents or using a sequence like Conway-Guy. I hope, by discarding the "sequence generator" constraint, I can find numbers much closer together than such sequences provide.
For any value of N, it is readily possible to generate up to Floor(Log2(N))-1 numbers (which we'll call the set "S") such that:
All members of S are less than or equal to N, and
No two distinct subsets of S have the same sum, and
All members of S are within a factor of two of each other.
Your suspicions were correct in that S would not be in any sense extensible (you could not add more members to it)
Method:
For N, find T = 2^P , where T is the highest power of two that is less than or equal to N. That is:
P = Floor( Log2(N) ), and
T = 2^P
Then the members of S can be generated as:
for( i=0 to P-2 ): S(i) = 2^i + 2^(P-1)
Or, to put it another way, S(i) = 2^i, for 0<= i < P-1
This makes for a total of P-1 (or Floor(Log2(N))-1) members. Can two distinct subsets of S ever sum to the same number? No:
Proof
Let's consider any two subsets of S: U and V, which are distinct (that is, they have no members in common). Then the sum of U is:
Sum(U) = O(U)*(T/2) + Sum(2^i| S(i):U)
Where
O(U) is the Order of the set U (how many elements it has),
"S(i):U" means "S(i) is an element of U", and
"|" is the conditioning operator (means "given that.." or "where.."),
So, putting the last two together, Sum(2^i| S(i):U) just means "the sum of all of the powers of two that are elements of U" (remembering that S(i) = 2^i)).
And likewise, the sum of V is:
Sum(V) = O(V)*(2^(P-1)) + Sum(2^i| S(i):V)
Now because U and V are distinct: Sum(2^i| S(i):U) can never be equal, because no two sums of distinct powers of two can ever be equal.
Also, because Sum(2^i; 0 <= i < P-1) = 2^(P-1)-1), these sums of the powers of two must always be less than 2^P-1. This means that the sums of U and V could only be equal if:
O(U)*(2^(P-1)) = O(V)*(2^(P-1))
or
O(U) = O(V)
That is, if U and V have the same number of elements, so that the first terms will be equal (because the second terms can never be as large as any differences in the first terms).
In such a case (O(U) = (O(V)) the first terms are equal, so Sum(U) would equal Sum(V) IFF their second terms (the binary sums) are also equal. However, we already know that they can never be equal, therefore, it can never be true that Sum(U) = Sum(V).
It seems like another way of phrasing the problem is to make sure that the previous terms never sum to the current term. If that's never the case, you'll never have two sums that add up to the same.
Ex: 2, 3, 6, 12, 24, 48, 96, ...
Summing to any single element {i} takes 1 more than the sum of the previous terms, and summing to any multi-element set {i,j} takes more than the sum of previous elements to i and previous elements to j.
More mathematically: (i-1), i, 2i, 4i, 8i, ... 2^n i Should work for any i, n.
The only way this doesn't work is if you're allowed to choose the same number twice in your subset (if that's the case, you should specify it in the problem). But that brings up the issue that Sum{i} = Sum{i} for any number, so that seems like an issue.
Related
Definition
Set P={e1,e2,...,en},P has n different elements,enumerated as ei's in it.
Set I={e1',e2',...,en'},I has at least one element that is similar to some element of P.The number of elements in I need not be equal to the number of elements in P.
Each I has a weight Q associated with it, and that describes the cost to use it .Q>0
You have to help me in designing an algorithm, that takes a set P as input, and some (say k of them) I sets, denoted by I1,I2,. . . , Ik, and exactly k, Q values, denoted by Q1,Q2,. . . ,Qk. Q1 denots the cost to use set I1, and so on.
You have to choose some I's, say I1,I2,. . . , such that when they all are unioned together, they produce set P' and P is a subset of that.
Notice that once you find a selection of I's, it has a cost associated with it.
You also have to make sure that this cost is as MINIMUM as possible.
Input
input one Set P
input a list of Set I,IList={I1,I2,...In}
input a list of Set Q,QList={Q1,Q2,...Qn}
Ix Qx are corresponding one by one.
Output
P' = Ia union Ib...union In
P' ⊂ P
Make the Qa+Qb...+Qn be the min value.
Also mention the Time and Space Complexity of your algorithm
Sample Input
P={a,b,c}
I1={a,x,y,z} Q1=0.7
I2={b,c,x} Q2=1
I3={b,x,y,z} Q3=2
I4={c,y} Q4=3
I5={a,b,c,y} Q5=9
Sample Output
P1 = I1 U I2 COST=Q1+Q2=1.7
P2 = I1 U I3 U I4 COST=Q1+Q3+Q4=5.7
P3 = I5 COST=Q5=9
And:P⊂P1,P⊂P2,P⊂P3
The P COST : 1.7<5.7<9
And then what we want is:
P1 = I1 U I2 COST=Q1+Q2=1.7
Here is some suggestion to simplify the problem.
We first duplicate all the I sets, and lets call them I1', I2', . . .
Now, first job that we should do is to remove the unwanted elements from duplicated I' sets. Here unwanted means the elements which will not contribute towards the main set P.
We discard all those I' sets which do not have even a single element of P.
Now suppose P has some n elements in it, we now know definitely that I' sets are nothing but subsets of the main set, and every subset has a cost Qi associated with it.
We just have to pick some subsets such that they together cover the main set.
Subject to the minimum cost.
We will denote the main set and subsets using bit based notation.
If the set P has n elements in it, we will have n bits in the representation.
So the main set will be denoted by <1,1,...1> (n 1's).
And it's subsets will be denoted by bitset, having some 1's absent from the bitset of main set. Because I's are also subsets, they will also have some binary representation denoting the subset they are representing.
To solve the problem efficiently, let's make an assumption that there is so much of memory available, that if the bitset is treated as a number in binary, we can index the bitsets, to some memory location in constant time.
This means that, if we have, suppose n = 4, all the subsets can be represented
by different values from 0 to 15 (see their binary representation from 0000(empty set) to 1111(main set), when element at position i of main array is present in a subset we put a 1 at that position in the bitset). And similarly when n is larger.
Now, having the bitset based notation for the set, the Union of two sets denoted by bitset b1 and b2 will be denoted by b1|b2. where | is bitwise OR operation.
Of course, we will not require so many memory locations, as not all the subsets of parent set will be available as I's.
Algorithm :
The algorithmic idea used here is bitset based Dynamic Programming.
Assume we have a big array, namely COST, where COST[j] represents the cost to have the subset, represented by bitset notation of j.
To start with the algorithm, we first put the cost to choose given subsets (in terms of I's), in their respective indices in COST array, and at all the other locations we put a very large value, say INF.
What we have to do is, to fill the array appropriately, and then once it is filled properly, we will get the answer to minimum cost by looking at the value COST[k] where k has all bits set, in binary representation.
Now we will focus on how to fill the array properly.
This is rather easy task, we will iterate the COST array, K no. of times where K is the no. of I'-sets we have.
For every I's set, let's call it's binary representation BI'.
we OR the bit representation of BI' and current index(idx), and what we get is the new set which is the UNION of the set represented by current index, and BI', let's call this new set as S' and it's final binary representation as BS'.
We will look at the COST[BS'], and if we see that this COST is larger than COST[BI'] + COST[idx], we will update the value at the COST[BS'].
In similar way we proceed, and at the end of the run, we get the minimum cost at COST[BP], where BP is the bitset for P.
In order to track the participating I's, who actually contributed in the formation of P, we can take a note, while updating any index.
TIME COMPLEXITY : O(2^n * K), where K is the no. of I sets, and n is the no. of elements in P.
Space Complexity : O(2^n)
NOTE : Because of the assumption, that the bit-representation are directly indexable, the solution may not be very much feasible for large values of n and k.
i have a range
R = {0, ..., N}
and i like to get K elements which have a sum equal to S, but the elements should be selected randomly.
So an easy brute force method would be to determine all element combinations containing K numbers resulting in S and picking one of the combinations by random.
I am trying to think about a recursive solution where a random number is selected and then the problem reduces to find (K-1) random numbers with sum equal to (S - K0) but this need not yield in a solution.
Is there a better approach?
A sample would be:
R = {0,1,2,3,4,5}, S = 5, K = 2
Solutions: randomly pick one of {{1,4};{2,3};{0.5}}
In general, if K is big (then N also), and S not too little, it is unpredictable, because, there are two many combinations.
Brute force: try every combinations. You are sure to find a solution, if there exists one, but if there are more than, say, 1 Md, or somewhat, it it almost impossible to list them all.
Your algorithm:
To choose at random, your algorithm is ok: take one number at random, then another, ...
But you make an assumption: there exists a solution with the numbers you pick: you dont know.
So what ? if statistically there exist many solutions, you could find it like that, perhaps, or perhaps not.
Some trails:
1 Use S/K
If every numbers < S/K, it is impossible.
if every numbers > S/K, it is impossible.
So lets assume that there are numbers < S/K, and other > S/K
2 keep only numbers < S, very interesting if S is little.
3 idea: If S is big, and numbers little, you have chance that there exist many combinations.
idea of algorithm
1 take one number N1 at random
2 if N1 < S/K, take another one N2 > S/K
3 calculate N1+N2: if < 2.S/K take another one N3> S/K, if not
4 iterate at each step: if sum < n S/K take another one > S/K, if not
5 you can have better precision, by replacing S/K by (S-sum N1,N2,...)/(K-n)
If at one step you dont can not find any number, backtrack
hope it helps
I would start with Dirichlet distribution (https://en.wikipedia.org/wiki/Dirichlet_distribution). Using it, you could sample uniformly in (0..1) distributed random numbers Xi, such that SumiXi = 1.
For S <= N, it is easy to see that sampling beyond S is useless and should be rejected outright.
So, combining with acceptance/rejection, something along the lines
Divide interval [0...1] into S (or S+1 if 0 is allowed) equal bins.
Sample K numbers from Dirichlet distribution.
Map sampled numbers to bin index, so you have now sampled integers which are
all below or equal S and have sum equal to S.
If all integers are distinct, accept the sampling, otherwise reject the sampling and go to step 2
I need an algorithm for this problem:
Given a set of n natural numbers x1,x2,...,xn, a number S and k. Form the sum of k numbers picked from the set (a number can be pick many times) with sum S.
Stated differently: List every possible combination for S with Bounds: n<=256, x<=1000, k<=32
E.g.
problem instance: {1,2,5,9,11,12,14,15}, S=30, k=3
There are 4 possible combinations
S=1+14+15, 2+14+14, 5+11+15, 9+9+12.
With these bounds, it is unfeasible to use brute force but I think of dynamic programming is a good approach.
The scheme is: Table t, with t[m,v] = number of combinations of sum v formed by m numbers.
1. Initialize t[1,x(i)], for every i.
2. Then use formula t[m,v]=Sum(t[m-1,v-x(i)], every i satisfied v-x(i)>0), 2<=m<=k.
3. After obtaining t[k,S], I can trace back to find all the combinations.
The dilemma is that t[m,v] can be increase by duplicate commutative combinations e.g., t[2,16]=2 due to 16=15+1 and 1+15. Furthermore, the final result f[3,30] is large, due to 1+14+15, 1+15+14, ...,2+14+14,14+2+14,...
How to get rid of symmetric permutations? Thanks in advance.
You can get rid of permutations by imposing an ordering on the way you pick elements of x. Make your table a triple t[m, v, n] = number of combinations of sum v formed by m numbers from x1..xn. Now observe t[m, v, n] = t[m, v, n-1] + t[m-1, v-x_n, n]. This solves the permutation problem by only generating summands in reverse order from their appearance in x. So for instance it'll generate 15+14+1 and 14+14+2 but never 14+15+1.
(You probably don't need to fill out the whole table, so you should probably compute lazily; in fact, a memoized recursive function is probably what you want here.)
I saw this post: Finding even numbers in an array and I was thinking about how you could do it without feedback. Here's what I mean.
Given an array of length n containing at most e even numbers and a
function isEven that returns true if the input is even and false
otherwise, write a function that prints all the even numbers in the
array using the fewest number of calls to isEven.
The answer on the post was to use a binary search, which is neat since it doesn't mean the array has to be in order. The number of times you have to check if a number is even is e log n instead if n because you do a binary search (log n) to find one even number each time (e times).
But that idea means that you divide the array in half, test for evenness, then decide which half to keep based on the result.
My question is whether or not you can beat n calls on a fixed testing scheme where you check all the numbers you want for evenness without knowing the outcome, and then figure out where the even numbers are after you've done all the tests based on the results. So I guess it's no-feedback or blind or some term like that.
I was thinking about this for a while and couldn't come up with anything. The binary search idea doesn't work at all with this constraint, but maybe something else does? Even getting down to n/2 calls instead of n (yes, I know they are the same big-O) would be good.
The technical term for "no-feedback or blind" is "non-adaptive". O(e log n) calls still suffice, but the algorithm is rather more involved.
Instead of testing the evenness of products, we're going to test the evenness of sums. Let E ≠ F be distinct subsets of {1, …, n}. If we have one array x1, …, xn with even numbers at positions E and another array y1, …, yn with even numbers at positions F, how many subsets J of {1, …, n} satisfy
(∑i in J xi) mod 2 ≠ (∑i in J yi) mod 2?
The answer is 2n-1. Let i be an index such that xi mod 2 ≠ yi mod 2. Let S be a subset of {1, …, i - 1, i + 1, … n}. Either J = S is a solution or J = S union {i} is a solution, but not both.
For every possible outcome E, we need to make calls that eliminate every other possible outcome F. Suppose we make 2e log n calls at random. For each pair E ≠ F, the probability that we still cannot distinguish E from F is (2n-1/2n)2e log n = n-2e, because there are 2n possible calls and only 2n-1 fail to distinguish. There are at most ne + 1 choices of E and thus at most (ne + 1)ne/2 pairs. By a union bound, the probability that there exists some indistinguishable pair is at most n-2e(ne + 1)ne/2 < 1 (assuming we're looking at an interesting case where e ≥ 1 and n ≥ 2), so there exists a sequence of 2e log n calls that does the job.
Note that, while I've used randomness to show that a good sequence of calls exists, the resulting algorithm is deterministic (and, of course, non-adaptive, because we chose that sequence without knowledge of the outcomes).
You can use the Chinese Remainder Theorem to do this. I'm going to change your notation a bit.
Suppose you have N numbers of which at most E are even. Choose a sequence of distinct prime powers q1,q2,...,qk such that their product is at least N^E, i.e.
qi = pi^ei
where pi is prime and ei > 0 is an integer and
q1 * q2 * ... * qk >= N^E
Now make a bunch of 0-1 matrices. Let Mi be the qi x N matrix where the entry in row r and column c has a 1 if c = r mod qi and a 0 otherwise. For example, if qi = 3^2, then row 2 has ones in columns 2, 11, 20, ... 2 + 9j and 0 elsewhere.
Now stack these matrices vertically to get a Q x N matrix M, where Q = q1 + q2 + ... + qk. The rows of M tell you which numbers to multiply together (the nonzero positions). This gives a total of Q products that you need to test for evenness. Call each row a "trial", and say that a "trial involves j" if the jth column of that row is nonempty. The theorem you need is the following:
THEOREM: The number in position j is even if and only if all trials involving j are even.
So you do a total of Q trials and then look at the results. If you choose the prime powers intelligently, then Q should be significantly smaller than N. There are asymptotic results that show you can always get Q on the order of
(2E log N)^2 / 2log(2E log N)
This theorem is actually a corollary of the Chinese Remainder Theorem. The only place that I've seen this used is in Combinatorial Group Testing. Apparently the problem originally arose when testing soldiers coming back from WWII for syphilis.
The problem you are facing is a form of group testing, type of a problem with the objective of reducing the cost of identifying certain elements of a set (up to d elements of a set of N elements).
As you've already stated, there are two basic principles via which the testing may be carried out:
Non-adaptive Group Testing, where all the tests to be performed are decided a priori.
Adaptive Group Testing, where we perform several tests, basing each test on the outcome of previous tests. Obviously, adaptive testing has a potential to reduce the cost, compared to non-adaptive testing.
Theoretical bounds for both principles have been studied, and are available in this Wiki article, or this paper.
For adaptive testing, the upper bound is O(d*log(N)) (as already described in this answer).
For non-adaptive testing, it can be shown that the upper bound is O(d*d/log(d)*log(N)), which is obviously larger than the upper bound for adaptive testing by a factor of d/log(d).
This upper bound for non-adaptive testing comes from an algorithm which uses disjunct matrices: matrices of dimension T x N ("number of tests" x "number of elements"), where each item can be either true (if an element was included in a test), or false (if it wasn't), with a property that any subset of d columns must differ from all other columns by at least a single row (test inclusion). This allows linear time of decoding (there are also "d-separable" matrices where fewer test are needed, but the time complexity for their decoding is exponential and not computationaly feasible).
Conclusion:
My question is whether or not you can beat n calls on a fixed testing scheme [...]
For such a scheme and a sufficiently large value of N, a disjunct matrix can be constructed which would have less than K * [d*d/log(d)*log(N)] rows. So, for large values of N, yes, you can beat it.
The underlying question (challenge) is kind of silly. If the binary search answer is acceptable (where it sums sub arrays and sends them to IsEven) then I can think of a way to do it with E or less calls to IsEven (assuming the numbers are integers of course).
JavaScript to demonstrate
// sort the array by only the first bit of the number
A.sort(function(x,y) { return (x & 1) - (y & 1); });
// all of the evens will be at the beginning
for(var i=0; i < E && i < A.length; i++) {
if(IsEven(A[i]))
Print(A[i]);
else
break;
}
Not exactly a solution, but just few thoughts.
It is easy to see that if a solution exists for array length n that takes less than n tests, then for any array length m > n it is easy to see that there is always a solution with less than m tests. So, if you have a solution for n = 2 or 3 or 4, then the problem is solved.
You can split the array into pairs of numbers and for each pair: if the sum is odd, then exactly one of them is even, otherwise if one of the numbers is even, then both of them are even. This way for each pair it takes either one or two tests. Best case:n/2 tests, worse case:n tests, if even and odd numbers are chosen with equal probability, then: 3n/4 tests.
My hunch is there is no solution with less than n tests. Not sure how to prove it.
UPDATE: The second solution can be extended in the following way.
Check if the sum of two numbers is even. If odd, then exactly one of them is even. Otherwise label the set as "homogeneous set of size 2". Take two "homogenous set"s of same size n. Pick one number from each set and check if their sum is even. If it is even, combine these two sets to a "homogeneous set of size 2n". Otherwise, it implies that one of those sets purely consists of even numbers and the other one purely odd numbers.
Best case:n/2 tests. Average case: 3*n/2. Worst case is still n. Worst case exists only when all the numbers are even or all the numbers are odd.
If we can add and multiply array elements, then we can compute every Boolean function (up to complementation) on the low-order bits. Simulate a circuit that encodes the positions of the even numbers as a number from 0 to nC0 + nC1 + ... + nCe - 1 represented in binary and use calls to isEven to read off the bits.
Number of calls used: within 1 of the information-theoretic optimum.
See also fully homomorphic encryption.
We've got some nonnegative numbers. We want to find the pair with maximum gcd. actually this maximum is more important than the pair!
For example if we have:
2 4 5 15
gcd(2,4)=2
gcd(2,5)=1
gcd(2,15)=1
gcd(4,5)=1
gcd(4,15)=1
gcd(5,15)=5
The answer is 5.
You can use the Euclidean Algorithm to find the GCD of two numbers.
while (b != 0)
{
int m = a % b;
a = b;
b = m;
}
return a;
If you want an alternative to the obvious algorithm, then assuming your numbers are in a bounded range, and you have plenty of memory, you can beat O(N^2) time, N being the number of values:
Create an array of a small integer type, indexes 1 to the max input. O(1)
For each value, increment the count of every element of the index which is a factor of the number (make sure you don't wraparound). O(N).
Starting at the end of the array, scan back until you find a value >= 2. O(1)
That tells you the max gcd, but doesn't tell you which pair produced it. For your example input, the computed array looks like this:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
4 2 1 1 2 0 0 0 0 0 0 0 0 0 1
I don't know whether this is actually any faster for the inputs you have to handle. The constant factors involved are large: the bound on your values and the time to factorise a value within that bound.
You don't have to factorise each value - you could use memoisation and/or a pregenerated list of primes. Which gives me the idea that if you are memoising the factorisation, you don't need the array:
Create an empty set of int, and a best-so-far value 1.
For each input integer:
if it's less than or equal to best-so-far, continue.
check whether it's in the set. If so, best-so-far = max(best-so-far, this-value), continue. If not:
add it to the set
repeat for all of its factors (larger than best-so-far).
Add/lookup in a set could be O(log N), although it depends what data structure you use. Each value has O(f(k)) factors, where k is the max value and I can't remember what the function f is...
The reason that you're finished with a value as soon as you encounter it in the set is that you've found a number which is a common factor of two input values. If you keep factorising, you'll only find smaller such numbers, which are not interesting.
I'm not quite sure what the best way is to repeat for the larger factors. I think in practice you might have to strike a balance: you don't want to do them quite in decreasing order because it's awkward to generate ordered factors, but you also don't want to actually find all the factors.
Even in the realms of O(N^2), you might be able to beat the use of the Euclidean algorithm:
Fully factorise each number, storing it as a sequence of exponents of primes (so for example 2 is {1}, 4 is {2}, 5 is {0, 0, 1}, 15 is {0, 1, 1}). Then you can calculate gcd(a,b) by taking the min value at each index and multiplying them back out. No idea whether this is faster than Euclid on average, but it might be. Obviously it uses a load more memory.
The optimisations I can think of is
1) start with the two biggest numbers since they are likely to have most prime factors and thus likely to have the most shared prime factors (and thus the highest GCD).
2) When calculating the GCDs of other pairs you can stop your Euclidean algorithm loop if you get below your current greatest GCD.
Off the top of my head I can't think of a way that you can work out the greatest GCD of a pair without trying to work out each pair individually (and optimise a bit as above).
Disclaimer: I've never looked at this problem before and the above is off the top of my head. There may be better ways and I may be wrong. I'm happy to discuss my thoughts in more length if anybody wants. :)
There is no O(n log n) solution to this problem in general. In fact, the worst case is O(n^2) in the number of items in the list. Consider the following set of numbers:
2^20 3^13 5^9 7^2*11^4 7^4*11^3
Only the GCD of the last two is greater than 1, but the only way to know that from looking at the GCDs is to try out every pair and notice that one of them is greater than 1.
So you're stuck with the boring brute-force try-every-pair approach, perhaps with a couple of clever optimizations to avoid doing needless work when you've already found a large GCD (while making sure that you don't miss anything).
With some constraints, e.g the numbers in the array are within a given range, say 1-1e7, it is doable in O(NlogN) / O(MAX * logMAX), where MAX is the maximum possible value in A.
Inspired from the sieve algorithm, and came across it in a Hackerrank Challenge -- there it is done for two arrays. Check their editorial.
find min(A) and max(A) - O(N)
create a binary mask, to mark which elements of A appear in the given range, for O(1) lookup; O(N) to build; O(MAX_RANGE) storage.
for every number a in the range (min(A), max(A)):
for aa = a; aa < max(A); aa += a:
if aa in A, increment a counter for aa, and compare it to current max_gcd, if counter >= 2 (i.e, you have two numbers divisible by aa);
store top two candidates for each GCD candidate.
could also ignore elements which are less than current max_gcd;
Previous answer:
Still O(N^2) -- sort the array; should eliminate some of the unnecessary comparisons;
max_gcd = 1
# assuming you want pairs of distinct elements.
sort(a) # assume in place
for ii = n - 1: -1 : 0 do
if a[ii] <= max_gcd
break
for jj = ii - 1 : -1 :0 do
if a[jj] <= max_gcd
break
current_gcd = GCD(a[ii], a[jj])
if current_gcd > max_gcd:
max_gcd = current_gcd
This should save some unnecessary computation.
There is a solution that would take O(n):
Let our numbers be a_i. First, calculate m=a_0*a_1*a_2*.... For each number a_i, calculate gcd(m/a_i, a_i). The number you are looking for is the maximum of these values.
I haven't proved that this is always true, but in your example, it works:
m=2*4*5*15=600,
max(gcd(m/2,2), gcd(m/4,4), gcd(m/5,5), gcd(m/15,15))=max(2, 2, 5, 5)=5
NOTE: This is not correct. If the number a_i has a factor p_j repeated twice, and if two other numbers also contain this factor, p_j, then you get the incorrect result p_j^2 insted of p_j. For example, for the set 3, 5, 15, 25, you get 25 as the answer instead of 5.
However, you can still use this to quickly filter out numbers. For example, in the above case, once you determine the 25, you can first do the exhaustive search for a_3=25 with gcd(a_3, a_i) to find the real maximum, 5, then filter out gcd(m/a_i, a_i), i!=3 which are less than or equal to 5 (in the example above, this filters out all others).
Added for clarification and justification:
To see why this should work, note that gcd(a_i, a_j) divides gcd(m/a_i, a_i) for all j!=i.
Let's call gcd(m/a_i, a_i) as g_i, and max(gcd(a_i, a_j),j=1..n, j!=i) as r_i. What I say above is g_i=x_i*r_i, and x_i is an integer. It is obvious that r_i <= g_i, so in n gcd operations, we get an upper bound for r_i for all i.
The above claim is not very obvious. Let's examine it a bit deeper to see why it is true: the gcd of a_i and a_j is the product of all prime factors that appear in both a_i and a_j (by definition). Now, multiply a_j with another number, b. The gcd of a_i and b*a_j is either equal to gcd(a_i, a_j), or is a multiple of it, because b*a_j contains all prime factors of a_j, and some more prime factors contributed by b, which may also be included in the factorization of a_i. In fact, gcd(a_i, b*a_j)=gcd(a_i/gcd(a_i, a_j), b)*gcd(a_i, a_j), I think. But I can't see a way to make use of this. :)
Anyhow, in our construction, m/a_i is simply a shortcut to calculate the product of all a_j, where j=1..1, j!=i. As a result, gcd(m/a_i, a_i) contains all gcd(a_i, a_j) as a factor. So, obviously, the maximum of these individual gcd results will divide g_i.
Now, the largest g_i is of particular interest to us: it is either the maximum gcd itself (if x_i is 1), or a good candidate for being one. To do that, we do another n-1 gcd operations, and calculate r_i explicitly. Then, we drop all g_j less than or equal to r_i as candidates. If we don't have any other candidate left, we are done. If not, we pick up the next largest g_k, and calculate r_k. If r_k <= r_i, we drop g_k, and repeat with another g_k'. If r_k > r_i, we filter out remaining g_j <= r_k, and repeat.
I think it is possible to construct a number set that will make this algorithm run in O(n^2) (if we fail to filter out anything), but on random number sets, I think it will quickly get rid of large chunks of candidates.
pseudocode
function getGcdMax(array[])
arrayUB=upperbound(array)
if (arrayUB<1)
error
pointerA=0
pointerB=1
gcdMax=0
do
gcdMax=MAX(gcdMax,gcd(array[pointera],array[pointerb]))
pointerB++
if (pointerB>arrayUB)
pointerA++
pointerB=pointerA+1
until (pointerB>arrayUB)
return gcdMax